Formulas we need (uniformly accelerated motion):
h = vo*t - (1/2)gt^2
vf^2 - vo^2 = 2gd
Let H,t be the height of tower.
The first stone thrown upward reaches its maximum height,
h,max = vo^2 / 2g
h,max = 15^2 / 19.6
h,max = 11.4796 meters (above the tower)
and the time needed to reach its maximum height,
h = vo*t - (1/2)gt^2
11.4796 = 15*t - 4.9*t^2
t = 1.531 seconds
It will be at this position after 2 seconds:
h = 15*2 - (1/2)*9.8*2^2
h = 10.4 meters above the tower.
At this point, we can say that the first stone is moving downwards (towards the ground) because it exceeded the time for it to reach the maximum height.
And its velocity at this point is,
vf^2 - vo^2 = 2gd
vf^2- 15^2 = -2*9.8*10.4
vf = 4.6
The equation for this setup would therefore be:
h = vo*t - (1/2)gt^2
H,t + 10.4 = 4.6*t - 4.9*t^2
For the second stone dropped from the tower (note that it started from rest, i.e. vo = 0),
h = vo*t - (1/2)gt^2
H,t = -4.9t^2
Solving the two equations,
-4.9t^2 + 10.4 = 4.6*t - 4.9*t^2
10.4 = 4.6t
t = 2.26 s
After 2.26 seconds the second stone is dropped, it reaches the ground. Finally, the height of tower is,
H,t = -4.9*2.26^2
H,t = 25.04 meters (absolute value)
I guess there is a shorter way of doing this though I'm not sure how. ;u;
And I'm also not sure of my answer. If there are corrections, feel free to correct me. Still, I hope this helps~ :)
A stone is thrown vertically upward from the top of the tower with a velocity of 15m / s. Two seconds later a second stone is dropped from the top of the tower.if both the stone strike the ground simultaneously find the height of the tower
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1 answer